Right triangle upper A upper B upper C is to the left of right triangle upper D upper E upper F. Sides upper A upper B and upper D upper E are marked at their centers by single tick marks. Squares are drawn in the interior angles by upper B and upper E.

Question

What theorem proves that △ABC≅△DEF
?

Option # 1: ASA

Option # 2: HL

Option # 3: SAS

Option # 4: SSA

(1 point)
△ABC≅△DEF
by Option #
.

Answer 1

To determine which theorem proves that triangle \( \triangle ABC \) is congruent to triangle \( \triangle DEF \), we can analyze the information given:

1. Both triangles are right triangles since they are specifically labeled as such.
2. The sides \( AB \) and \( DE \) are marked at their centers by single tick marks, indicating that these sides are equal in length.
3. The presence of squares in the interior angles at points \( B \) and \( E \) indicates that both angles \( \angle ABC \) and \( \angle DEF \) are right angles (90 degrees).

Given this information, we can use the Right Triangle Congruence Theorem:

• Right Angle Hypotenuse Leg (HL): This theorem states that if we have two right triangles with congruent hypotenuses and one corresponding leg that is also congruent, then the triangles are congruent.

In this case:

• The right angles \( \angle ABC \) and \( \angle DEF \) are congruent.
• The sides \( AB \) and \( DE \) are marked as congruent.
• The sides opposite those right angles (the hypotenuses \( AC \) and \( DF \)) are also congruent by default because they are the sides opposite the right angles.

Thus, we can conclude:

\[ \triangle ABC \cong \triangle DEF \text{ by Option # } \text{HL.} \]

Final Answer: \( \triangle ABC \cong \triangle DEF \) by Option # 2.